The correct option is D n=4
Given
(1+ax)n=1+8x+24x2+… ⋯(i)
Also,
(1+ax)n= nC0(1)n(ax)0+ nC1(1)n−1(ax)1+ nC2(1)n−2(ax)2⋯ ⋯(ii)
Comparing the coefficients of x and x2, we get
⇒ nC1a=8 and nC2a2=24⇒na=8 and n(n−1)2a2=24
Now,
n(n−1)2a2=24⇒na(na−a)=48⇒8(8−a)=48 (∵na=8)⇒a=2⇒n=4